Solve 3t^2+12t-15

$$3 { t }^{ 2 } +12t-15$$

Factor

$$3\left(t-1\right)\left(t+5\right)$$

Solution Steps

Factor out 3.

$$3\left(t^{2}+4t-5\right)$$

Consider t^{2}+4t-5. Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt-5. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-5\right)=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=5$$

Rewrite t^{2}+4t-5 as \left(t^{2}-t\right)+\left(5t-5\right).

$$\left(t^{2}-t\right)+\left(5t-5\right)$$

Factor out t in the first and 5 in the second group.

$$t\left(t-1\right)+5\left(t-1\right)$$

Factor out common term t-1 by using distributive property.

$$\left(t-1\right)\left(t+5\right)$$

Rewrite the complete factored expression.

$$3\left(t-1\right)\left(t+5\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$3t^{2}+12t-15=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$t=\frac{-12±\sqrt{12^{2}-4\times 3\left(-15\right)}}{2\times 3}$$

Square 12.

$$t=\frac{-12±\sqrt{144-4\times 3\left(-15\right)}}{2\times 3}$$

Multiply -4 times 3.

$$t=\frac{-12±\sqrt{144-12\left(-15\right)}}{2\times 3}$$

Multiply -12 times -15.

$$t=\frac{-12±\sqrt{144+180}}{2\times 3}$$

Add 144 to 180.

$$t=\frac{-12±\sqrt{324}}{2\times 3}$$

Take the square root of 324.

$$t=\frac{-12±18}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{-12±18}{6}$$

Now solve the equation t=\frac{-12±18}{6} when ± is plus. Add -12 to 18.

$$t=\frac{6}{6}$$

Divide 6 by 6.

$$t=1$$

Now solve the equation t=\frac{-12±18}{6} when ± is minus. Subtract 18 from -12.

$$t=-\frac{30}{6}$$

Divide -30 by 6.

$$t=-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and -5 for x_{2}.

$$3t^{2}+12t-15=3\left(t-1\right)\left(t-\left(-5\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$3t^{2}+12t-15=3\left(t-1\right)\left(t+5\right)$$

Evaluate

$$3\left(t-1\right)\left(t+5\right)$$